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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
Thursday at 11:16 PM
0 replies
Solution needed ASAP
UglyScientist   4
N a minute ago by MathsII-enjoy
$ABC$ is acute triangle. $H$ is orthocenter, $M$ is the midpoint of $BC$, $L$ is the midpoint of smaller arc $BC$. Point $K$ is on $AH$ such that, $MK$ is perpendicular to $AL$. Prove that: $HMLK$ is paralelogram
4 replies
UglyScientist
an hour ago
MathsII-enjoy
a minute ago
Inequality
MathsII-enjoy   0
10 minutes ago
A interesting problem generalized :-D
0 replies
MathsII-enjoy
10 minutes ago
0 replies
Interesting inequalities
sqing   1
N 20 minutes ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ (a+b)^2 (a+c)^2=16abc. $ Prove that
$$ 2a+b+c\leq \frac{128}{27}$$$$ \frac{9}{2}a+b+c\leq \frac{864}{125}$$$$3a+b+c\leq 24\sqrt{3}-36$$$$5a+b+c\leq \frac{4(8\sqrt{6}-3)}{9}$$
1 reply
1 viewing
sqing
Yesterday at 2:35 PM
sqing
20 minutes ago
Geometry Problem
Euler_Gauss   0
24 minutes ago
Given that $D$ is the midpoint of $BC$, $DM$ bisects $\angle ADB$ and intersects $AB$ at $M$, $I$ is the incenter of $\triangle {}{}{}ABD$, $AT$ bisects $\angle BAC$ and intersects the circumcircle of \(\triangle {}{}ABC\) at $T$, $MS$ is parallel to $BC$ and intersects $AT$ at $S$. Prove that $\angle MIS + \angle BIT = \pi.$
0 replies
Euler_Gauss
24 minutes ago
0 replies
Angle AEB
Ecrin_eren   2
N an hour ago by sunken rock
In triangle ABC, the lengths |AB|, |BC|, and |CA| are proportional to 4, 5, and 6, respectively. Points D and E lie on segment [BC] such that the angles ∠BAD, ∠DAE, and ∠EAC are all equal. What is the measure of angle ∠AEB in degrees?

2 replies
Ecrin_eren
Yesterday at 9:26 AM
sunken rock
an hour ago
Where to check solutions??
math_gold_medalist28   0
an hour ago
I'm studying MONT by aditya khurmi and pathfinder by vikash tiwari...but the problem is there isn't given the solutions means the ans. So how can I be sure that my ans is correct or not ?Please help!!!
0 replies
math_gold_medalist28
an hour ago
0 replies
How many triangles
Ecrin_eren   1
N 3 hours ago by Ecrin_eren


"Inside a triangle, 2025 points are placed, and each point is connected to the vertices of the smallest triangle that contains it. In the final state, how many small triangles are formed?"


1 reply
Ecrin_eren
Yesterday at 11:55 AM
Ecrin_eren
3 hours ago
How many integer pairs
Ecrin_eren   3
N 3 hours ago by gasgous

"Let m and n be integers. How many different integer pairs (m, n) satisfy the equation m^3 - 3m^2n + 4n^3 = 44?"

3 replies
Ecrin_eren
Yesterday at 12:02 PM
gasgous
3 hours ago
Values of x
Ecrin_eren   4
N 3 hours ago by vanstraelen
Given 0 ≤ x < 2π, what is the difference between the largest and the smallest of the values of x
that satisfy the equation 5cosx + 2sin2x = 4 in radians?
4 replies
Ecrin_eren
Yesterday at 6:42 PM
vanstraelen
3 hours ago
All possible values of k
Ecrin_eren   3
N 4 hours ago by sqing


The roots of the polynomial
x³ - 2x² - 11x + k
are r₁, r₂, and r₃.

Given that
r₁ + 2r₂ + 3r₃ = 0,
what is the product of all possible values of k?

3 replies
Ecrin_eren
Yesterday at 8:42 AM
sqing
4 hours ago
How many pairs
Ecrin_eren   2
N 5 hours ago by Ecrin_eren


Let n be a natural number and p be a prime number. How many different pairs (n, p) satisfy the equation:

p + 2^p + 3 = n^2 ?



2 replies
Ecrin_eren
Yesterday at 3:08 PM
Ecrin_eren
5 hours ago
Inequalities
sqing   1
N 6 hours ago by sqing
Let $ a,b,c\geq 0 ,a+b+c =4. $ Prove that
$$2a +ab +ab^2c \leq\frac{63+5\sqrt 5}{8}$$$$2a +ab^2 +abc \leq \frac{4(68+5\sqrt {10})}{27}$$$$     2a +a^2b + a b^2c^3\leq \frac{4(50+11\sqrt {22})}{27}$$
1 reply
sqing
Today at 3:57 AM
sqing
6 hours ago
Inequalities
sqing   3
N Today at 7:55 AM by sqing
Let $ a,b \geq 0 $ and $ a-b+a^3-b^3=2  $.Prove that$$  a^2+ab+b^2 \geq 1 $$Let $ a,b \geq 0 $ and $ a+b+a^3+b^3=2  $.Prove that$$    a^2-ab+b^2  \leq 1 $$
3 replies
sqing
Today at 3:02 AM
sqing
Today at 7:55 AM
Range of a trigonometric function
Saucepan_man02   4
N Today at 6:52 AM by brownbear.bb
Find the range of the function: $f(x)=\frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2}$.
4 replies
Saucepan_man02
Apr 28, 2025
brownbear.bb
Today at 6:52 AM
Combo problem about regular polygons
NJAX   1
N Apr 27, 2025 by Titibuuu
Source: 1st TASIMO, Day1 Problem3
$Abdulqodir$ cut out $2024$ congruent regular $n-$gons from a sheet of paper and placed these $n-$gons on the table such that some parts of each of these $n-$gons may be covered by others. We say that a vertex of one of the afore-mentioned $n-$gons is $visible$ if it is not in the interior of another $n-$gon that is placed on top of it. For any $n>2$ determine the minimum possible number of visible vertices.

Proposed by David Hrushka, Slovakia
1 reply
NJAX
May 18, 2024
Titibuuu
Apr 27, 2025
Combo problem about regular polygons
G H J
G H BBookmark kLocked kLocked NReply
Source: 1st TASIMO, Day1 Problem3
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NJAX
29 posts
#1 • 2 Y
Y by dmusurmonov, Seungjun_Lee
$Abdulqodir$ cut out $2024$ congruent regular $n-$gons from a sheet of paper and placed these $n-$gons on the table such that some parts of each of these $n-$gons may be covered by others. We say that a vertex of one of the afore-mentioned $n-$gons is $visible$ if it is not in the interior of another $n-$gon that is placed on top of it. For any $n>2$ determine the minimum possible number of visible vertices.

Proposed by David Hrushka, Slovakia
This post has been edited 3 times. Last edited by NJAX, May 18, 2024, 9:41 AM
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Titibuuu
43 posts
#2 • 2 Y
Y by B1t, sami1618
Suppose first that $n \geq 3$ is even, write $n = 2k$, $k \geq 2$, and consider an $n$-gon $X$ which is not covered by any other (there must clearly be at least one, take any if there are more than one) and an $n$-gon $Y$ which has this property after removing $X$ from the table. Since any diagonal connecting a pair of opposite vertices in $Y$ is the longest among all segments contained in $Y$, and hence also in $X$, no pair of opposite vertices of $Y$ can be covered by the interior of $X$.

It follows that there are at least $k$ visible vertices of $Y$ which together with $2k$ visible vertices from $X$ make a lower bound $3k$.

If $n$ is odd, i.e., $n = 2k + 1$ for some $k \geq 1$, we choose $X$ and $Y$ in the same way and moreover pick $Z$ being ``on top'' after both $X$ and $Y$ are removed. Then we argue analogously as above (using the fact that the diagonal connecting a vertex $V$ of an $(2k+1)$-gon $G$ with any endpoint of the opposite side to $V$ in $G$ realizes the longest segment contained in $G$) to conclude that at least $k + 1$ vertices from $Y$ are visible and at least one vertex of $Z$ must be visible. Thus, we get a lower bound of
\[
(2k+1) + (k+1) + 1 = 3k+3.
\]
The following constructions prove that those lower bounds are achievable for any $n \geq 3$.

First, consider the case $n = 2k$ again. Take three copies $P, Q, R$ of a regular $2k$-gon, place $R$ with its side $AB$ going horizontally and being the bottom one, $Q$ precisely on top of $R$, and $P$ precisely on top of $Q$. Then rotate $Q$ around $A$ and $P$ around $B$, one clockwise, the other counterclockwise, both by equally large angles so that the segment $AB$ is not covered by $P$ and $Q$ anymore and the opposite vertex of $A$ in $Q$ and the opposite vertex of $B$ in $P$ meet.

It is easy to see that only $A$ and $B$ are now visible among the vertices of $R$. Shifting finally $P$ and $Q$ down by distances $d$ and $2d$ for some $d>0$ small enough makes no vertices of $R$ visible and precisely $k$ vertices of $Q$ visible. The remaining $2019$ $2k$-gons can be placed identically as $R$ (under it) which makes in total $2k+k=3k$ visible vertices.

Secondly, let $n = 2k+1$. Analogously, place four copies of a $(2k+1)$-gon on the table, one precisely on top of the previous ones, one common side lying horizontally and below. Then observe that shifting the top layer to the right, the second one to the left, then the first one down, then the second one also down (but less), and finally the third one up, with all distances chosen properly (and after hiding $2018$ more copies of our $(2k+1)$-gon under the fourth layer) we get $(2k+1) + (k+1) + 1 = 3k+3$ visible vertices.
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